1480. Running Sum of 1d Array.java

/*
Given an array nums. We define a running sum of an array as runningSum[i] = sum(nums[0]…nums[i]).

Return the running sum of nums.

Example 1:

Input: nums = [1,2,3,4]
Output: [1,3,6,10]
Explanation: Running sum is obtained as follows: [1, 1+2, 1+2+3, 1+2+3+4].
Example 2:

Input: nums = [1,1,1,1,1]
Output: [1,2,3,4,5]
Explanation: Running sum is obtained as follows: [1, 1+1, 1+1+1, 1+1+1+1, 1+1+1+1+1].
Example 3:

Input: nums = [3,1,2,10,1]
Output: [3,4,6,16,17]
*/

//Solution
class Solution {
    public int[] runningSum(int[] nums) {
        int[] sum = new int[nums.length];
        for(int i = 0;i < nums.length;i++){
            if(i >= 1){
                sum[i] = sum[i-1] + nums[i];
            }else{
                sum[i] = nums[i];
            }
        }
        return sum;
        
    }
}